Classical mechanics (find trajectory and kinetic energy)

[NanoMath]

Homework Statement

Given the force $\vec{ F }(x) = (-12x + 6) \vec{i}$ ; find kinetic energy $T$ at the point $x=2$ and trajectory of a particle $\vec{r}(t)$, given that $\vec{r}(t=0)=\vec{0}$ and $\dot{\vec{r}}(t=0)=\vec{0}$ .

3. The Attempt at a Solution
Since $\nabla\times \vec{F} = \vec{0}$ we can find potential function $\vec{F}(x) = -\nabla U(x)$ , therefore $U(x)= 6x²-6x + C$ . From here I am not sure how to proceed.
Second idea is to solve differential equation but I'm not sure how to do it because force is a function of $x$ and not $t$ .
$$ m\ddot{x}=-12x+6$$ $$\ddot{x}+\frac{1}{m}12x-\frac{1}{m}6=0$$ let's suppose $m=1$ for simplicity, then we get $$ \ddot{x}+12x-6=0$$

 

[haruspex]

Homework Statement

Given the force $\vec{ F }(x) = (-12x + 6) \vec{i}$ ; find kinetic energy $T$ at the point $x=2$ and trajectory of a particle $\vec{r}(t)$, given that $\vec{r}(t=0)=\vec{0}$ and $\dot{\vec{r}}(t=0)=\vec{0}$ .

3. The Attempt at a Solution
Since $\nabla\times \vec{F} = \vec{0}$ we can find potential function $\vec{F}(x) = -\nabla U(x)$ , therefore $U(x)= 6x²-6x + C$ . From here I am not sure how to proceed.
Second idea is to solve differential equation but I'm not sure how to do it because force is a function of $x$ and not $t$ .
$$ m\ddot{x}=-12x+6$$ $$\ddot{x}+\frac{1}{m}12x-\frac{1}{m}6=0$$ let's suppose $m=1$ for simplicity, then we get $$ \ddot{x}+12x-6=0$$

With a simple change of variable you can eliminate the constant 6 from the differential equation and obtain a very well known form. Think springs.
By the way, this belongs in Introductory Physics homework.